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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2011, Article ID 875649, 15 pages
doi:10.1155/2011/875649
Research Article
Nonsquareness and Locally Uniform
Nonsquareness in Orlicz-Bochner Function
Spaces Endowed with Luxemburg Norm
Shaoqiang Shang,
1
Yunan Cui,
2
and Yongqiang Fu
1
1
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2
Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China
Correspondence should be addressed to Shaoqiang Shang,
Received 5 July 2010; Accepted 12 February 2011
Academic Editor: Nikolaos Papageorgiou
Copyright q 2011 Shaoqiang Shang et al. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Criteria for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces
equipped with Luxemburg norm are given. We also prove that, in Orlicz-Bochner function
spaces generated by locally uniform nonsquare Banach space, nonsquareness and locally uniform
nonsquareness are equivalent.
1. Introduction
A lot of nonsquareness concepts in Banach spaces are known see 1. Nonsquareness are
important notions in geometry of Banach space. One of reasons is that these properties are
strongly related to the fixed point property see 2. The criteria for nonsquareness and
locally uniform nonsquareness in the classical Orlicz function spaces have been given in 3, 4
already. However, because of the complicated structure of Orlicz-Bochner function spaces
equipped with the Luxemburg norm, the criteria for nonsquareness and locally uniform
nonsquareness of them have not been found yet. The aim of this paper is to give criteria
for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces
equipped with Luxemburg norm.
Let X, · be a real Banach space. SX and BX denote the unit sphere and unit
ball, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach
space X is said to be nonsquare if for any x, y ∈ SX we have min{xy/2, x−y/2} <
1. A Banach space X is said to be uniformly nonsquare if there exists δ>0 such that for any
x, y ∈ SX,min{x y/2, x − y/2
} < 1 − δ. A Banach space X is said to be locally
uniformly nonsquare if for any x ∈ SX, there exists δ
x
> 0 such that min{x y/2, x −
y/2} < 1 − δ
x
, where y ∈ SX.
2 Journal of Inequalities and Applications
Let R be set of real numbers. A function M : R → R
is called an N-function if M is
convex, even, M00, Mu > 0 u
/
0 and lim
u → 0
Mu/u0, and lim
u → 0
Mu/u
∞.
Let T, Σ,μ be a nonatomic measurable space. p denotes right derivative of M.
Moreover, for a given Banach space X, ·, we denote by X
T
the set of all strongly μ-
measurable function from T to X, and for each u ∈ X
T
, we define the modular of u by
ρ
M
u
G
M
u
t
dt.
1.1
Put
L
M
u
t
∈ X
T
:
G
M
λu
t
dt < ∞ for some λ>0
.
1.2
The linear set L
M
endowed with the Luxemburg norm
u
inf
λ>0:ρ
M
u
λ
≤ 1
1.3
is a Banach space. We say that an Orlicz function M satisfies condition Δ
2
M ∈ Δ
2
if there
exist K>2andu
0
≥ 0 such that
M
2u
≤ KM
u
u ≥ u
0
. 1.4
First let us recall a known result that will be used in the further part of the paper.
Lemma 1.1 see 3. Suppose M ∈ Δ
2
.Then
ρ
M
u
n
−→ 0 ⇐⇒
u
n
−→ 0,ρ
M
u
n
−→ 1 ⇐⇒
u
n
−→ 1
n −→ ∞
. 1.5
2. Main Results
Theorem 2.1. L
M
is nonsquare if and only if
a M ∈ Δ
2
;
b X is nonsquare.
In order to prove the theorem, we give a lemma.
Lemma 2.2. If X is nonsquare, then for any x, y
/
0, we have
x
y
− min
x y
,
x − y
> 0. 2.1
Journal of Inequalities and Applications 3
Proof.
Case 1. If x < y, then
x y
≤
x
x
y
· y
1 −
x
y
·
y
<
x
x
y
−
x
x
y
2.2
or
x − y
<
x
y
. 2.3
Case 2. If x≥y, then
x y
≤
y
x
· x y
x
y
− 1
·
y
<
y
y
x
−
y
x
y
2.4
or
x − y
<
x
y
. 2.5
This implies x y−min{x y, x − y} > 0. This completes the proof.
Proof of Theorem 2.1. a Necessity. Suppose that M/∈ Δ
2
, then there exist u ∈ SL
M
and δ>0
such that ρ
M
u1 − δ<1. Pick c>0 such that E {t ∈ T : ut≤c} is not a null set. Since
M/∈ Δ
2
, there exist sequence {r
n
}
∞
n1
and disjont subsets {E
n
}
∞
n1
of E such that
r
n
> 2nc, M
1
1
n
r
n
> 2
n
M
1
1
2n
r
n
, 2
n
M
1
1
2n
r
n
μE
n
2
−n
δ.
2.6
Therefore, if we define v
∞
n1
r
n
χ
E
n
, then for any l>1, we have
ρ
M
lv
∞
n1
ρ
M
lr
n
χ
E
n
≥
∞
nm
ρ
M
1
1
n
r
n
χ
E
n
>
∞
nm
2
n
ρ
M
1
1
2n
r
n
χ
E
n
≥
∞
nm
2
n
M
1
1
2n
r
n
μE
n
∞
nm
2
n
· 2
−n
δ ∞,
ρ
M
v
∞
n1
M
r
n
μE
n
<
∞
n1
M
r
n
c
μE
n
<
∞
n1
M
1
1
2n
r
n
μE
n
δ.
2.7
4 Journal of Inequalities and Applications
This yields
v
1,ρ
M
u ± v
≤ ρ
M
u
∞
n1
M
r
n
c
μE
n
1 − δ δ 1.
2.8
Hence, u ± v≤1. But u v u − v≤2u 2, and we deduce that u v u − v 1.
Moreover, we have 1/2u v1/2u − v 1and1/2u v − 1/2u − v} 1,
a contradiction with nonsquareness of L
M
.
If b is not true, then there exist x, y ∈ SX such that x y 1/2x y
1/2x − y.Pickα>0 such that
T
Mαdt 1. Put
u
t
α · x · χ
T
t
,v
t
α · y · χ
T
t
. 2.9
Then we have
ρ
M
u
T
M
αx
dt
T
M
α
dt 1,
ρ
M
v
T
M
αy
dt
T
M
α
dt 1.
2.10
It is easy to see u, v ∈ SL
M
. We know that
u
t
v
t
2
α ·
x y
2
· χ
T
t
,
u
t
− v
t
2
α ·
x − y
2
· χ
T
t
.
2.11
Hence, we have
ρ
M
u v
2
T
M
α ·
x y
2
dt
T
M
α
dt 1,
ρ
M
u − v
2
T
M
α ·
x − y
2
dt
T
M
α
dt 1.
2.12
It is easy to see 1/2u v, 1/2u − v ∈ SL
M
, a contradiction!
Sufficiency. Suppose that there exists u, v ∈ SL
M
such that
u
v
1
2
u v
1
2
u − v
1. 2.13
We will derive a contradiction for each of the following two cases.
Journal of Inequalities and Applications 5
Case 1. μ{t ∈ T : ut
/
0}∩{t ∈ T : vt
/
0}0. Let G {t ∈ T : ut
/
0}. Hence, we
have
1
2
ρ
M
u
1
2
ρ
M
v
1
2
G
M
u
t
dt
1
2
T\G
M
v
t
dt
1
2
G
M
u
t
v
t
dt
1
2
T\G
M
u
t
v
t
dt
>
G
M
1
2
u
t
v
t
dt
T\G
M
1
2
u
t
v
t
dt
T
M
1
2
u
t
v
t
dt
ρ
M
1
2
u v
.
2.14
Since M ∈ Δ
2
, we have ρ
M
uρ
M
v1. Hence, ρ
M
u v/2 < 1. This implies u
v/2 < 1, a contradiction!
Case 2. μ{t ∈ T : ut
/
0}∩{t ∈ T : vt
/
0} > 0. By Lemma 2.2, without loss of
generality, we may assume that there exists T
1
⊂{t ∈ T : ut
/
0}∩{t ∈ T : vt
/
0} such
that ut vt > utvt, t ∈ T
1
and μT
1
> 0. Therefore,
1
2
ρ
M
u
1
2
ρ
M
v
1
2
T
M
u
t
dt
1
2
T
M
v
t
dt
T
1
2
M
u
t
1
2
M
v
t
dt
≥
T
1
M
1
2
u
t
1
2
v
t
dt
T\T
1
M
1
2
u
t
1
2
v
t
dt
>
T
1
M
1
2
u
t
v
t
dt
T\T
1
M
1
2
u
t
v
t
dt
ρ
M
u v
2
.
2.15
Since M ∈ Δ
2
, we have ρ
M
uρ
M
v1. Hence, ρ
M
u v/2 < 1. This implies
u v/2 < 1, a contradiction!
Theorem 2.3. L
M
is locally uniformly nonsquare if and only if
a M ∈ Δ
2
;
b X is locally uniformly nonsquare.
6 Journal of Inequalities and Applications
In order to prove the theorem, we give a lemma.
Lemma 2.4. If X is locally uniformly nonsquare, then
a For any x
/
0, r
1
≥ r
2
> 0, we have
inf
y
/
0
x
y
− min
x y
,
x − y
: x ∈ X, r
2
≤
y
≤ r
1
> 0
2.16
b If x
n
→ x,thenlim
n →∞
δx
n
δx,where
δ
x
inf
y
/
0
x
y
− min
x y
,
x − y
: x ∈ X, r
2
≤
y
≤ r
1
.
2.17
Proof. a Since X is locally uniformly nonsquare, we have η
x
> 0andη
λx
λη
x
, where λ>0
and
η
x
inf
y
x
y
− min
x y
,
x − y
:
x
y
> 0
.
2.18
In fact, since X is locally uniformly nonsquare, we have
η
x
inf
y
x
y
− min
x y
,
x − y
:
x
y
> 0
x
· inf
y
2 − min
x
x
y
y
,
x
x
−
y
y
:
x
y
> 0
> 0,
η
λx
inf
y
λx
y
− min
λx y
,
λx − y
:
λx
y
> 0
λ · inf
y
x
1
λ
y
− min
x
1
λ
y
,
x −
1
λ
y
:
x
1
λ
y
> 0
λ · inf
y
x
y
− min
x y
,
x − y
:
x
y
> 0
λ · η
x
.
2.19
Journal of Inequalities and Applications 7
Case 1. If x≥y, then
x y
≤
1 −
y
x
x y
y
x
x
≤
1 −
y
x
x
y
y
x
x
≤
x
−
y
y
y
− η
y/xx
≤
x
y
− η
r
2
/xx
2.20
or
x − y
≤
x
y
− η
r
2
/xx
. 2.21
Case 2. If x < y, then
x y
≤
x
y
· y x
1 −
x
y
·
y
≤
x
x
− η
x
y
−
x
x
y
− η
x
≤
x
y
− η
x
2.22
or
x − y
≤
x
y
− η
x
. 2.23
Therefore, we get, the following inequality
inf
y
x
y
− min
x y
,
x − y
: x ∈ X
≥ min
η
r
2
/xx
,η
x
> 0
2.24
holds.
b1 Suppose that lim sup
n →∞
δx
n
>δx, where x
n
→ x n →∞. Then there exist
a>0 and subsequence {n} of {n}, such that δx
n
− δx ≥ a. By definition of δx, there exist
y
0
∈ X such that
x
y
0
− min
x y
0
,
x − y
0
<δ
x
a
8
,r
1
≤
y
0
≤ r
2
.
2.25
We will derive a contradiction for each of the following two cases.
8 Journal of Inequalities and Applications
Case 1. x y
0
x − y
0
. Since x
n
→ x n →∞, there exists n
0
such that x
n
0
− x <a/8.
Therefore,
x
n
0
y
0
−
x
n
0
y
0
≤
x
x
n
0
− x
y
0
−
x
n
0
y
0
≤
x
x
n
0
− x
y
0
−
x y
0
−
x
n
0
− x
x
y
0
−
x y
0
2
x
n
0
− x
≤ δ
x
a
8
2
x
n
0
− x
<δ
x
a
8
2 ·
a
8
δ
x
3
8
a.
2.26
This implies δx
n
0
≤ δx3/8a, a contradiction!
Case 2. x − y
0
/
x y
0
. Without loss of generality, we may assume x − y
0
> x y
0
r,
where r>0. Since x
n
→ x n →∞, there exists n
0
such that x
n
0
−x < min{1/8a, 1/8r}.
Therefore, we have
x
n
0
− y
0
x − y
0
x
n
0
− x
≥
x − y
0
−
x
n
0
− x
≥
x − y
0
−
1
8
r,
x
n
0
y
0
x y
0
x
n
0
− x
≤
x y
0
x
n
0
− x
≤
x y
0
1
8
r.
2.27
This implies
x
n
0
− y
0
≥
x − y
0
−
1
8
r ≥
x y
0
r −
1
8
r ≥
x y
0
1
8
r ≥
x
n
0
y
0
.
2.28
Similarly, we have
x
n
0
y
0
−
x
n
0
y
0
≤ δ
x
3
8
a.
2.29
Therefore, we have
x
n
0
y
0
− min
x
n
0
y
0
,
x
n
0
− y
0
≤ δ
x
3
8
a.
2.30
This implies δx
n
0
≤ δx3/8a, a contradiction! Hence, lim sup
n →∞
δx
n
≤ δx.
Journal of Inequalities and Applications 9
b2 Suppose that lim inf
n →∞
δx
n
<δx, where x
n
→ x n →∞. Then there exist
b>0 and subsequence {n} of {n}, such that δx − δx
n
≥ b. Since x
n
→ x n →∞, then
there exist n
0
∈ N such that x
n
0
− x
n
< 1/8b, whenever n ≥ n
0
. By definition of δx
n
0
,
there exist y
0
∈ X such that
x
n
0
y
0
− min
x
n
0
y
0
,
x
n
0
− y
0
<δ
x
n
0
b
8
,r
2
≤
y
0
≤ r
1
.
2.31
Therefore, we have
x
n
y
0
− min
x
n
y
0
,
x
n
− y
0
x
n
0
− x
n
0
x
n
y
0
− min
x
n
0
− x
n
0
x
n
y
0
,
x
n
0
− x
n
0
x
n
− y
0
≤
x
n
0
1
8
b
y
0
− min
x
n
0
y
0
,
x
n
0
− y
0
1
8
b
x
n
0
y
0
− min
x
n
0
y
0
,
x
n
0
− y
0
1
4
b
<δ
x
n
0
1
8
b
1
4
b
<δ
x
n
0
3
8
b
2.32
whenever n ≥ n
0
. Since x
n
→ x n →∞, there exists n
1
>n
0
such that |ηx − ηx
n
1
| <
1/8b, where
η
x
x
y
0
− min
x y
0
,
x − y
0
. 2.33
Hence, we have
η
x
n
1
>η
x
−
1
8
b ≥ δ
x
−
1
8
b ≥ δ
x
n
0
b −
1
8
b δ
x
n
0
7
8
b.
2.34
This implies
x
n
1
y
0
− min
x
n
1
y
0
,
x
n
1
− y
0
≥ δ
x
n
0
7
8
b,
2.35
which contradict 2.32. Hence, lim inf
n →∞
δx
n
≥ δx.
Combing b1 with b2, we get lim
n →∞
δx
n
δx. This completes the proof.
10 Journal of Inequalities and Applications
Proof of Theorem 2.3. Necessity.ByTheorem 2.1, M ∈ Δ
2
.Ifb is not true, then there exist
x ∈ SX, {y
n
}
∞
n1
⊂ SX such that 1/2x y
n
→1and1/2x − y
n
→1asn →∞.
Pick α>0 such that
T
Mαdt 1. Put
u
t
α · x · χ
T
t
,v
n
t
α · y
n
· χ
T
t
. 2.36
Then we have
ρ
M
u
T
M
αx
dt
T
M
α
dt 1,
ρ
M
v
n
T
M
αy
n
dt
T
M
α
dt 1.
2.37
It is easy to see u, v
n
∈ SL
M
. We know that
u
t
v
n
t
2
α ·
x y
n
2
· χ
T
t
,
u
t
− v
n
t
2
α ·
x − y
n
2
· χ
T
t
.
2.38
Moreover, we have Mα·xy
n
/2 ≤ Mα, Mα·x−y
n
/2 ≤ Mα. By the dominated
convergence theorem, we have
lim
n →∞
T
M
α ·
x y
n
2
dt
T
lim
n →∞
M
α ·
x y
n
2
dt
T
M
α
dt 1,
lim
n →∞
T
M
α ·
x − y
n
2
dt
T
lim
n →∞
M
α ·
x − y
n
2
dt
T
M
α
dt 1.
2.39
It is easy to see ρ
M
1/2u v
n
→ 1, ρ
M
1/2u− v
n
→ 1asn →∞.ByLemma 1.1,we
have 1/2u v
n
→1and1/2u − v
n
→1asn →∞, a contradiction with locally
uniform nonsquareness of L
M
.
Sufficiency. Suppose that there exist u ∈ SL
M
, {v
n
}
∞
n1
⊂ SL
M
such that 1/2u
v
n
→1, 1/2u − v
n
→1asn →∞. We will derive a contradiction for each of the
following two cases.
Case 1. There exist ε
0
> 0, σ
0
> 0 such that μG
n
>ε
0
, where G
n
{t ∈ T : v
n
t≥σ
0
}.Put
H
n
t ∈ T : σ
0
≤
v
n
t
≤ M
−1
4
ε
0
. 2.40
We have
1
T
M
v
n
t
dt ≥
G
n
\H
n
M
v
n
t
dt ≥
4
ε
0
· μ
G
n
\ H
n
.
2.41
Journal of Inequalities and Applications 11
This implies μG
n
\ H
n
≤ 1/4ε
0
. Hence, μH
n
> 1/2ε
0
. We define a function
η
t
inf
y
/
0
u
t
y
− min
u
t
y
,
u
t
− y
: σ
0
≤
y
≤ M
−1
4
ε
0
2.42
on T
0
, where T
0
{t ∈ T : ut
/
0}.ByLemma 2.4, we have ηt > 0 μ-a.e on T
0
.Let
h
n
t → ut μ-a.e on T
0
, where h
n
is simple function. Hence,
η
n
t
inf
y
/
0
h
n
t
y
−min
h
n
t
y
,
h
n
t
−y
: σ
0
≤
y
≤M
−1
4
ε
0
2.43
is μ-measurable. By Lemma 2.4, we have η
n
t → ηt μ-a.e on T
0
. Then ηt is μ-measurable.
Using
T ⊃
∞
i1
t ∈ T
0
:
1
i 1
<η
t
≤
1
i
,
2.44
we get that there exists η
0
> 0 such that μH < 1/8ε
0
, where
H
t ∈ T
0
: η
t
≤ 2η
0
. 2.45
Let E
n
H
n
\ H, E
1
n
H
n
∩{t ∈ T : ut
/
0} \ H, E
2
n
H
n
∩{t ∈ T : ut 0} \ H.It
is easy to see E
n
E
1
n
∪ E
2
n
, E
1
n
∩ E
2
n
φ and μE
n
≥ 3/8ε
0
.Ift ∈ E
1
n
,byLemma 2.4, we have
u
t
v
n
t
− min
{
u
t
v
n
t
,
u
t
− v
n
t
}
≥ η
t
≥ 2η
0
. 2.46
Without loss of generality, we may assume that there exists F
1
n
⊂ E
1
n
such that
μF
1
n
≥
1
2
μE
1
n
,
u
t
v
n
t
−
u
t
v
n
t
≥ 2η
0
t ∈ F
1
n
.
2.47
Moreover, for any u ≥ v>0, we have
1
2
M
u
− M
1
2
u
−
1
2
M
v
− M
1
2
v
1
2
u
0
p
t
dt −
1/2u
0
p
t
dt −
1
2
v
0
p
t
dt
1/2v
0
p
t
dt
1
2
u
0
p
t
dt −
1
2
v
0
p
t
dt
−
1/2u
0
p
t
dt −
1/2v
0
p
t
dt
1
2
u
v
p
t
dt −
1/2u
1/2v
p
t
dt
≥
1/2u1/2v
v
p
t
dt −
1/2u
1/2v
p
t
dt ≥ 0.
2.48
12 Journal of Inequalities and Applications
Hence, if t ∈ E
2
n
, then
1
2
M
v
n
t
1
2
M
v
n
t
≥
1
2
M
σ
0
− M
σ
0
2
> 0.
2.49
Let F
n
F
1
n
∪ E
2
n
. Then μF
n
≥ 1/8ε
0
. Therefore,
1
2
ρ
M
u
1
2
ρ
M
v
n
− ρ
M
u v
n
2
1
2
T
M
u
t
dt
1
2
T
M
v
n
t
dt −
T
M
u
t
v
n
t
2
dt
T
1
2
M
u
t
1
2
M
v
n
t
− M
u
t
v
n
t
2
dt
≥
F
1
n
1
2
M
u
t
1
2
M
v
n
t
− M
u
t
v
n
t
2
dt
E
2
n
1
2
M
u
t
1
2
M
v
n
t
− M
u
t
v
n
t
2
dt
≥
F
1
n
M
1
2
u
t
1
2
v
n
t
− M
u
t
v
n
t
2
dt
E
2
n
1
2
M
u
t
1
2
M
v
n
t
− M
u
t
v
n
t
2
dt
≥
F
1
n
M
u
t
v
n
t
2
η
0
− M
u
t
v
n
t
2
dt
E
2
n
1
2
M
v
n
t
− M
v
n
t
2
dt
≥
F
1
n
M
η
0
dt
E
2
n
1
2
M
σ
0
− M
σ
0
2
dt
≥
F
n
min
M
η
0
,
1
2
M
σ
0
− M
σ
0
2
dt
min
M
η
0
,
1
2
M
σ
0
− M
σ
0
2
· μF
n
≥ min
M
η
0
,
1
2
M
σ
0
− M
σ
0
2
·
1
8
ε
0
.
2.50
By Lemma 1.1, we have ρ
M
uρ
M
v
n
1, ρ
M
u v
n
/2 → 1asn →∞.Thisisin
contradiction with 1/2ρ
M
u1/2ρ
M
v
n
− ρ
M
u v
n
/2 ≥ min{Mη
0
, 1/2Mσ
0
−
Mσ
0
/2}·1/8ε
0
.
Journal of Inequalities and Applications 13
Case 2. For any ε>0, σ>0, there exists N such that μ{t ∈ T : v
n
t≥σ} <εwhenever
n>N. By the Riesz theorem, without loss of generality, we may assume that v
n
t → 0 μ−
a.e on T.Using
{
t ∈ T :
u
t
/
0
}
⊃
∞
n1
t ∈ T :
1
n 1
<
u
t
≤
1
n
,
2.51
we get that there exist r>0 such that μT
1
< 1/8μT
0
, where
T
1
{
t ∈ T :0<
u
t
<d
}
,T
0
{
t ∈ T :
u
t
/
0
}
. 2.52
Since M is N-function, we can choose 0 <h<dsuch that 1/2Md1/2Mh − Md
h/2 > 0. Since v
n
t → 0μ− a.e on T, by the Egorov theorem, there exists N
1
such that
v
n
t <h,t∈ F whenever n>N
1
, where F ⊂ T, μT \ F < 1/8μT
0
. Next, we will prove
that if u
1
≥ u
2
≥ v
2
≥ v
1
> 0, then
1
2
M
u
1
1
2
M
v
1
− M
u
1
v
1
2
≥
1
2
M
u
2
1
2
M
v
2
− M
u
2
v
2
2
.
2.53
In fact, we have
1
2
M
u
1
1
2
M
v
1
− M
u
1
v
1
2
−
1
2
M
u
1
1
2
M
v
2
− M
u
1
v
2
2
1
2
M
v
1
− M
u
1
v
1
2
−
1
2
M
v
2
− M
u
1
v
2
2
1
2
v
1
0
p
t
dt −
u
1
v
1
/2
0
p
t
dt −
1
2
v
2
0
p
t
dt −
u
1
v
2
/2
0
p
t
dt
1
2
v
1
0
p
t
dt −
1
2
v
2
0
p
t
dt
u
1
v
2
/2
0
p
t
dt −
u
1
v
1
/2
0
p
t
dt
−
1
2
v
2
v
1
p
t
dt
u
1
v
2
/2
u
1
v
1
/2
p
t
dt
≥
u
1
v
2
/2
u
1
v
1
/2
p
t
dt −
v
2
v
1
v
2
/2
p
t
dt ≥ 0.
2.54
14 Journal of Inequalities and Applications
Moreover, we have
1
2
M
u
1
− M
u
1
v
2
2
−
1
2
M
u
2
− M
u
2
v
2
2
1
2
u
1
0
p
t
dt −
1
2
u
2
0
p
t
dt
−
u
1
v
2
/2
0
p
t
dt −
u
2
v
2
/2
0
p
t
dt
1
2
u
1
u
2
p
t
dt −
u
1
v
2
/2
u
2
v
2
/2
p
t
dt
≥
u
1
u
2
/2
u
2
p
t
dt −
u
1
v
2
/2
u
2
v
2
/2
p
t
dt ≥ 0.
2.55
By 2.54 and 2.55, we have
1
2
M
u
1
1
2
M
v
1
− M
u
1
v
1
2
≥
1
2
M
u
2
1
2
M
v
2
− M
u
2
v
2
2
.
2.56
This shows that if t ∈ T
2
T
0
\ T
1
∪ T \ F, then
M
u
t
M
v
n
t
− 2M
u
t
v
n
t
2
≥ M
d
M
h
− 2M
d h
2
.
2.57
It is easy to see μT
2
≥ 1/4μT
0
. Therefore,
1
2
ρ
M
u
1
2
ρ
M
v
n
− ρ
M
u v
n
2
1
2
T
M
u
t
dt
1
2
T
M
v
n
t
dt −
T
M
u
t
v
n
t
2
dt
T
1
2
M
u
t
1
2
M
v
n
t
− M
u
t
v
n
t
2
dt
≥
T
1
2
M
u
t
1
2
M
v
n
t
− M
u
t
v
n
t
2
dt
≥
T
2
1
2
M
u
t
1
2
M
v
n
t
− M
u
t
v
n
t
2
dt
≥
T
2
1
2
M
d
1
2
M
d
− M
d h
2
dt
≥
1
2
M
d
1
2
M
d
− M
d h
2
·
1
4
μT
0
,
2.58
for n large enough. By Lemma 1.1, we have ρ
M
uρ
M
v
n
1, ρ
M
uv
n
/2 → 1asn →∞,
which contradicts 1/2ρ
M
u1/2ρ
M
v
n
− ρ
M
u v
n
/2 ≥ 1/2Md1/2Md −
Md h/2 · 1/4μT
0
,forn large enough. This completes the proof.
Journal of Inequalities and Applications 15
Corollary 2.5. The following statements are equivalent:
a L
M
is locally uniformly nonsquare if and only if L
M
is nonsquare;
b X is locally uniformly nonsquare.
Acknowledgments
The authors would like to thank the anonymous referee for some suggestions to improve
the manuscript. This work was supported by China Natural Science Fund under Grant no.
11061022.
References
1 R. C. James, “Uniform nonsquare Banach space,” Annals of Mathematics, vol. 80, no. 3, pp. 542–550,
1964.
2 J. Garc
´
ıa-Falset, E. Llorens-Fuster, and E. M. Mazcu
˜
nan-Navarro, “Uniformly nonsquare Banach spaces
have the fixed point property for nonexpansive mappings,” Journal of Functional Analysis, vol. 233, no.
2, pp. 494–514, 2006.
3 S. T. Chen, “Geometry of Orlicz spaces,” Dissertationes Math, vol. 356, pp. 1–204, 1996.
4 C. X. Wu, T. F. Wang, S. T. Chen, and Y. W. Wang, Geometry Theory of Orlicz Spaces, H.I.T Print House,
Harbin, China, 1986.
